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pychu [463]
2 years ago
8

Consider the situation||: A child pulls a sled by a rope across the lawn at a constant speed. Of the forces listed, identify whi

ch act upon the sled.
Normal, Gravity, Applied, Friction, Tension, & Air Resistance
Physics
1 answer:
Mice21 [21]2 years ago
6 0

Answer:

Gravitational

Tension

Normal

Friction.

Explanation:

The forces acting on the sled are:

Tension: the tension from the rope, this is the force that "moves" the sled.

Friction: kinetic friction between the sled and the ground as the sled moves.

There are another two forces that also act on the sled, but that "has no effect"

Gravitational force: This force pulls the sled down, against the floor.

Normal force: This force "opposes" to the gravitational one, so they cancel each other.

These two forces cancel each other, so they have no direct impact on the movement of the sled. BUT, the friction force depends on the weight of the moving object, and the weight of the moving object depends on the gravitational force, so we need gravitational force in order to have friction force.

Then we can conclude that the forces acting on the sled are:

Gravitational

Tension

Normal

Friction.

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Physics 1102 Experiment 5 Pre Lab Name_________________________________ Instructor name _________________________ You must show
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Answer:

Explanation:

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3 years ago
Develop a hypothesis for why one of the two types of soup should indeed be rolling down faster than the other. This hypothesis s
Eva8 [605]

Answer:

Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.

incline angle = θ ; can's mass =  m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2

T1 = Inertia of Can with high viscosity soup

T2 = Inertia of Can with low viscosity soup

M1 rolling moment of Can 1

M2 rolling moment of Can 2

equation is given by

T1*α1 = M1   - (a)

T2*α2 = M2 - (b)

M1 = M2 = m*g*R*sin(θ). (c)

as assumed T1 > T2

from the three equation (a), (b) & (c)

the α2 > α1

Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.

7 0
3 years ago
A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan
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To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

\text{Initial Energy}=\text{Final Energy}

\frac{1}{2} mv^2=\frac{1}{2} kx^2

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

mv^2 = kx^2

v^2 = \frac{kx^2}{m}

v = \sqrt{\frac{kx^2}{m}}

Our values are,

m = 5.22*10^4kg

k = 4.58*10^5N/m

x = 32cm = 0.32m

Replacing our values we have,

v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}

v = 2.733*10^5m/s

Therefore the velocity is 2.733*10^5m/s

8 0
3 years ago
How does the gravitational force between two objects change if the distance
Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

So doubling the distance reduces the force 4 times.

7 0
3 years ago
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